Python:Extrema

This page discusses two different ways of getting Python to find the minimum of a function (versus a data set). Assuming the scipy.optimize module is loaded as opt, we will look at opt.fminbound and opt.fmin. The fminbound command can find a single independent value that will minimize a one-dimensional function over a specific domain. The fmin command can find a single array of values that will minimize a multi-dimensional function given some initial guess.

fminbound

The fminbound command can find a single value of a single variable that minimizes a function in a bounded interval. The function requires three arguments:

• The function to minimize,
• The left side of the boundary, and
• The right side of the boundary,

and, by default, returns the value of the variable that minimizes the function on the interval. If the kwarg full_output is True, then the function returns four pieces of information:

• The value of the variable that minimizes the function on the interval,
• The value of the function at the minimum,
• A flag that states if minimization worked (0) or not (1), and
• How many times the algorithm had to call the function to find the minimum value.

The function can be defined any way Python functions can be defined, including using a lambda function in the first argument. For example, to find the minimum value of \(\cos(x)\) for values of \(x\) between 0 and 6, the following will work:

def fun(x):
    return np.cos(x)

out1 = opt.fminbound(fun, 0, 6)
out4 = opt.fminbound(fun, 0, 6, full_output=True)

and the outputs are:

In [1]: out1
Out[1]: 3.1415926891518526

In [2]: out4
Out[2]: (3.1415926891518526, -0.9999999999999993, 0, 9)

Using lambda functions, you can skip the formal definition of the function and go straight for:

out1a = opt.fminbound(lambda v: np.cos(v), 0, 6)
out4a = opt.fminbound(lambda v: np.cos(v), 0, 6, full_output=True)

which produces the same values.

If there are multiple local minima, the algorithm may fail to reach the "most minimum" minimum. For instance, the result of:

def fun2(x):
    return np.cos(x)+x/20

test1 = opt.fminbound(fun2, 0, 40, full_output=True)

is

In [1]: test1
Out[1]: (9.374756357078237, -0.5300113625734499, 0, 12)

even though there is a more "more minimum" minimum in that domain. Specifically, note:

test2 = opt.fminbound(fun2, 0, 6, full_output=True)

yields:

In [2]: test2
Out[2]: (3.0915714695721217, -0.8441706279326544, 0, 9)

which is the actual minimum value of the function over the domain [0, 40].